IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VO L.23 (3 ) 2010 Dual No tions of Prime Modules I. M. Ali, R. I. Khalaf Departme nt of Mathematics, College of Education I bn-Al-Haitham, Unive rsity of Baghdad Abstract Let R be a commutative rin g with unity .M an R-M odule. M is called cop rime module (dual notion of prime module) if ann M =ann M /N for every p rop er submodule N of M In this p aper we study cop rime modules we give many basic p rop erties of this concept. Also we give many characterization of it under certain of module. Introduction Let R be a commutative rin g with unity . M an R-M odule. N is called p rime submodule of an R-modules M if N ≠ M and whenever rx  N such that r  R, x  M , then either x N or r  [N: M ], see [1], [2] .M is called a p rime modu le if ann M = ann N for every non-zero R R submodule N of M , see [3],[4] .It is clear that M is p rime iff <0> is Prime submodule of M . Yassemi.s. in [5], introduced the notion of second submodule (as dual notion of p rime submodule) as follows: N is second submodule of an R-modu le M if for every r  R, the homothety r* on N is either zero or surjective, where if M is an R-module and r R, then an R-endomorp hism r* is called homothety if r*(x) = rx for all x  M . M is second module if it is second submodule of M . Annin. S. in [6] introduced the notion of coprime modu les (as dual notion p rime modules) as follows: An R-modules M is called cop rime if ann M =ann M /N for every R R p rop er submodule N of M .Not that [N: M ] = ann M /N .Sp ecially a rin g R is coprime iff R is R R coprime R-module. Abuhilail J.in [7] in introduced a notion coprime submodules (as dual notion p rime submodules) as follows: N is called cop rime submodule of an R-module M if ann M =W R (M /N) = {a  R; the homothety r* on M /N is not surjective}, see [5].We notice that M is coprime module iff M is second module iff (0) is cop rime submodule. The main p urp ose of this p ap er is t o give basic p rop erties of coprime (second) modules and study the relationships between coprime modules and other modules . We show that a submodule of coprime module n eed not be cop rime modu le and we give certain conditions to make a submodu le of coprime module is coprime. M oreover we obtained the relationships between coprime modules and d ivisible modules, p rincipally injective and injective modules. Finally we invest igate the behavior of cop rime modu les under localization. Defini tion 1:[6] An R-module M is called coprime if for every p rop er submodule N of M , R ann N = R ann (M / N). IHJPAS IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VO L.23 (3 ) 2010 Sp ecially , a ring R is coprime iff R is a coprime R-module. Recall that a p rop er submodule N of an R-module M is called invariant if for each f  R nd (M ), f (N)  N. M is called ful ly invariant if every submodule of M is invariant, see [8]. .Wijay anti I.E in [9], gave the following characterization for coprime modules. The orem 2: [9] Let M be an R-module, then the following st atements are equivalent: 1. M is a cop rime R-module. 2. R ann M = R ann (M / N) for every p rop er invariant submodule N of M . Proof: (1)  (2) is obvious. (2)  (1). To p rove M is a cop rime R-module. Assume that R ann (M / N)  R ann M for some prop er submodule N of M . Let I = R ann (M / N). Then I  R ann M and I M  N. But I M is invariant, since for each f  R nd (M ), f (I M ) = I f (M )  I M . Hence I  R ann (M / I M ) = R ann M , which is a contradiction. ■ Not e that, st atement (2) in theorem (2) is used to define coprime modules in [27, Definition 1.3.1]. First , we give some remarks and examp les of coprime modules. Remarks and Examples 3: 1. Z as Z-module is not coprime. 2. Q as Z -module is cop rime module. 3. p Z  as a Z-module is a cop rime module [9]. 4. Every simple R-module is a coprime module, however the converse is not true for examp le: Q as Z -module is cop rime and not simple. Recall that an R-module M is called multiplication if for any submodule N of M , there exist an ideal I of R such that IM =N, equivalently for every submodule N of M , N= [N: M ]M, see [10]. 5. If M is a multip lication cop rime module, then M is simple, and hence M is p rime. Proof: Since M is a coprime R-module, then R ann M = [N R : M ] for all p rop er submodule N of M . Hence ( R ann M )M = [N R : M ]M . Thus (0) = [N R : M ]M . But M is a multiplication R-module, so [N R : M ]M = N, and hence N = (0). Thus M is a simp le R-module, and M is a prime R-module. ■ The condition "M is multiplication" can not be drop p ed from p revious remark for examp le: p Z  is a coprime Z-module, and by [2, Remark 1.1.3(11)] p Z  is not p rime and not multiplication. 6. If M is a cy clic coprime R-module, then M is simple and so p rime. 7. R is a cop rime ring if and only if R is a field. Proof: The proof follows by (Rem. and Ex. 2. (5), (4)). ■ 8. Zn is a cop rime Z-module iff n is a prime number. Proof: The proof follows directly by (Rem. and Ex. 3 (4), (5)). ■ 9. For any n, m  Z; n  m, the Z-module M = Z n Zm is not coprime. 10. Every vector sp ace M over a field R is a coprime R-module. IHJPAS IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VO L.23 (3 ) 2010 We have the following p rop osition: Proposi tion (4): Let M be an R-module. Consider the following st atements: 1. M is a cop rime R-module. 2. R ann M = [x R : M ] for every x  M such that (x) is a prop er submodule of M . 3. For every ideal I of R and for every x  M , (x) is a p rop er submodule of M such that I  ( )x  = (x), imp lies I = 0 or I M = 0. Then (1)  (2)  (3), and (2)  (1) if i R R i [ : ] [ R : ] i i x x       , where xi M and  is any index set. Proof: (1)  (2). It is clear. (2)  (3). Let I  ( )x  = (x) and assume I  0, then I M  (x), that is I  [x R : M ] and by (2) I  R ann M . Thus I M = (0). (3)  (2). Let r  [x R : M ]; x  M and (x) p rop er submodule of M . Then r  R ann ( )x  , that is (r) ( )x  = 0, hence by (3), either r = 0 or (r) M = 0. Thus r  R ann M . (2)  (1). Let N be a prop er submodule of M . Then N = R xi, xi  N. So that [N R : M ] = [R xi R : M ] = i R [ : ] i x N x    = R ann M . ■ The following p rop osition is a characterization of coprime module under the class of finitely generated (multiplication) modules. Proposi tion (5): Let M be a finitely generated (or multiplication) R-module, then M is a coprime R- module if and only if R ann M = [N R : M ], for every p rime submodule N of M . Proof: () It is clear. To p rove the converse, let W be a p rop er submodule of M . Since M is finitely generated (or multiplication) R-module, then by [15], [1] there exists a maximal submodule N of M (which is p rime by [30, Corollary 2.5, ch.1]) such that W  N M . Hence [W R : M ]  [N R : M ]. But [N R : M ] = R ann M by assump tion. Thus [W R : M ]  R ann M , and so R ann M = [W R : M ]. Recall that a submodule N of an R-module M is called second submodule if for every r  R, the homothety r* on N is either zero or surjective, where if M is an R-module and r R, then an R-endomorp hism r* is called homothety if r*(x) = rx for all x  M ; see [29, Definition 2.1. (b)]. M is second module if it is second submodule of M Remark (6): It is clear that N is a second submodule of an R-module iff for every rR, r≠0, either rNN or rN0. The following result is an interesting characterization of coprime modules. The orem (7): IHJPAS IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VO L.23 (3 ) 2010 Let M be an R-module, then M is coprime R-module iff M is a second R-module. Proof: () Let M be a cop rime R-module. Let r  R such that r  0. Sup p ose the homothety r* on M is not surjective, so r M  M . Let r M = N, then it is clear that r  [N R : M ]. Since M is a cop rime, then [N R : M ] = R ann M . Hence r  R ann M , that is r M = 0. Thus r * = 0. () To p rove R ann (M / N) = R ann M for every p rop er submodule N of M . Let r  [N R : M ], then r M  N  M . Since M is a second submodule, then the homothety r * on M is either zero or surjective. If r * is surjective, then r *(M ) = r M = M , imp lies M  N which is a contradiction. Thus r M = 0 and so r  R ann M . Therefore R ann   = R ann M . ■ The following result is an immediate consequence of Theorem (7) and [16, Rem. and Ex. 1.1.4 (3)). Note (8): If M is a cop rime R-module, then R ann M is a p rime ideal, and R / R ann M is an integral domain. So that we shall say that M is P-coprime if M is coprime with R ann M = P. A series of results follows by using theorem (7). Corollary (9): [28] Let M be an R-module, then M is coprime iff for every r  R, r  0, either r M = 0 or r M = M (i.e. M is a second module). Proof: It follows directly by Theorem (7) and Remark (6). ■ Corollary (10): Let M be an R-module, let I be an ideal of R such that I  R ann M . Thus M is a coprime R-module if and only if M is a cop rime R / I-module. Proof: It follows by Theorem (7) and [16, Rem. and Ex. (1.1.4(8)). ■ Corollary (11): Let M be an R-module. Then M is a coprime R-module iff M is a coprime R = R / R ann M -module. Proof: It follows directly by p revious corollary. ■ Corollary (12): If R is an integral domain, then Q(R) [the total quotient field of R] is a coprime R- module. Proof: Is obvious. Corollary (13): The homomorphic image of coprime R-module is coprime. Proof: It follows by Theorem (7) and [16.Rem. and Ex. (1.1.4(5)]. ■ Not e that,Wijayanti I.E. p roved that if M is a coprime R-module and N is an invariant submodule of M , then M / N is a coprime R-module, see [9, Prop . 1.3.8]. Hence the following corollary is a stronger result. Corollary (14): If M is a cop rime R-module, then M / W is a cop rime R-module. Proof: Let : M  M / W be the natural p rojection. Hence the result follows by p revious corollary . ■ Corollary (15): IHJPAS IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VO L.23 (3 ) 2010 Let M , W be two R-modules such that M  W, then M is coprime iff W is cop rime. Proof: It is immediate by corollary (14). ■ By considering (Rem. and Ex. 3 (1), (2)) a submodule of coprime R-module (second module) need not be cop rime R-module. However, in the following p rop osition, this is t rue under certain condition. Proposi tion (16): Let N be a non-zero p rop er submodule of an R-module M such that r M  N = r N, for every r  R, then M is P-cop rime iff N and M / N are P-cop rime R-modules. Proof: If M is a P-coprime R-module, then M is coprime with R ann M = P. Thus by corollary (14) M / N is coprime. Since P = R ann M = R ann (M / N), then M /N is P-cop rime. Now, to p rove N is a P-cop rime module. Since M is coprime, then for any r  R, r  0, either r M = 0 or r M = M . If r M = 0, then r M  N = 0. But r M  N = r N, t hen r N = 0. If r M = M , so r M  N = N. But r M  N = r N, hence r N = N. T hus N is second. To p rove N = P = R ann M . It is clear that R ann M  R ann N. Let r  R ann N, t hus r N = 0. If r M = 0, there is nothing to p rove. If r M = M , then r M  N = N. But r M  N = r N. Thus r N = N and so N = (0) which is a contradiction. Conversely , if N and M / N are P-cop rime. Then P = R ann N = R ann   and r N = N, r   =   for every r  P. T o p rove M is P-cop rime. It is clear R ann M  R ann N = R ann   = P. Let r  P, so r N = 0 and r M  N. Hence r M  N = r M , but r M  N = r N, so r M = r N = 0. Thus r  R ann M , hence P = R ann M . Let r  R ann M = P and let m  M , then m + N    = r   , hence m + N=r (m+N) for some m  M . Thus m – r m  N = r N, so m – r m = r n for some n  N and hence m = r (m + n)  r M . Thus M = r M for every r  P, and so M is P-cop rime. ■ Recall that a ring R is said to be regular (in sense of Von Neumann) if for each x  R, there exists a  R such that x = x 2  a, see [6]. Corollary (17): Let M be a module over a regular ring (in sense of Von Neumann) and let N be a submodule of M . Then M is a P-cop rime module iff N and M /N are P-cop rime R- modules. Proof: Since R is a regular ring, then for every r  R, r M  N = r N. Hence the result follows by p rop osition (16). ■ Now, we have the following result. Proposi tion (18): Let N be a finitely generated submodule of an R-module M such that R ann N = R ann M . If N is a coprime R-module, then M is a cop rime R-module. Proof: Let r  R and r  R ann M . Since N is a coprime module and R ann N = R ann M , r N = N. But N is a finitely generated submodule, so by [25, p .50], there exists r   R such that (1 – r r ) N = 0. Hence (1 – r r ) M = 0, and so M = r M . Therefore M is coprime. ■ Not e that, the condition R ann N = R ann M can not drop p ed from the previous p rop osition as the following examp le shows: IHJPAS IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VO L.23 (3 ) 2010 ( 2 ) in Z6 as Z-module is coprime module, and ann  ( 2 ) = 3Z  ann  (Z6) = 6Z. However Z6 is not coprime Z-module, see (Rem. and Ex. 3 (8)). Next, we can give the following p rop osition. Proposi tion (19): Let N be a finitely generated submodule of M and R ann M is a prime ideal. If N is a coprime module, then R ann M = [N R : M ]. Proof: Let r  [N R : M ], then r M  N. It is clear that r  R ann N or r  R ann N. If r  R ann N, then r N = 0 and hence r 2 M  r N = 0; that is r 2  R ann M . Since R ann M is p rime, then r  R ann M . If r  R ann N, t hen r N = N because N is a coprime R-module. On t he other hand, N is finitely generated, there exists r   R such that (1 – r r) N = 0, see [25, p.50]. It follows that r (1 – r r  ) M  (1 – r r  ) N = 0. Thus r (1 – r r  ) M = 0 and so r (1 – r r  )  R ann M , which implies either r  R ann M or (1 – r r  )  R ann M . If (1 – r r )  R ann M , then M = r M  N which is a contradiction. Thus r  R ann M and so [N R : M ] = R ann M . Recall that an R-module M is said to be Noetherian if every submodule of M is finitely generated, see [19, Prop . 6.2, p .75]. The following result is an immediate consequence of p rop osition (19). Corollary (20): Let M be a Noetherian R-module such that R ann M is a p rime ideal of R, if every submodule of M is a cop rime R-module, then M is coprime. Recall that a submodule N of an R- module M is called primary if for every rR, xM such that rxN, then either xN or r  R [ : ]  = {s  R; s n  R [ : ]  for some n  Z+}, see [20]. We know that every p rime submodule is p rimary; however the converse is not true in general. The following result shows that the two concepts are equivalent in the class of coprime modules. Proposi tion (21): Let M be a cop rime R-module, let N be a prop er submodule of M , then N is p rimary iff N is p rime. Proof: Let N be a primary submodule of M , since M is coprime, then R ann M = [N R : M ], so [N R : M ] is a prime ideal, hence by [14, Prop . 2.10, ch.1], we have N is p rime. ■ S.Yassem in [5] gave the following result. We give its p roof for the sake of comp leteness. Recall that a submodule N of an R-module M is called se condary submodul e if for each rR,the homothety r* on N is either surjective or nilpotent, where r* is nilpotent if there exist kZ+ such that (r*) =0,see[21]. Proposi tion (22): Let N be a submodule of an R-module M such that M is P-second (P- coprime), then N is P-secondary iff N is P-second (P-cop rime). Proof: () Since M is P-cop rime, R ann M = P. Sup p ose that N is P-secondary, then P = R ann  . Thus P = R ann  = R ann  . To p rove N is P-second. Since N is a submodule of IHJPAS IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VO L.23 (3 ) 2010 M , then P = R ann  R ann N, but R ann N  R ann  = P which is a p rime ideal. Thus by p rop osition (7) N is a second submodule with P = R ann N; that is N is P-second submodule. The converse is obvious by [15, Rem. and Ex. 1.2.2 (1)]. ■ Recall that if M is an R-module, then J(M) is the intersection of all maximal submodules of M (if exist) or J(M ) = M if M has no maximal submodule, see [12, Definition 9.1.2,p .214]. The following result ap p eared in [27]. However we give another p roof. Proposi tion (23): If M is a coprime R-module and J(M )  M , then R is a coprime ring, where R = R / R ann M . Proof: Since J(M )  M , then there exists a maximal submodule N of M . By [15, Prop o. 2.2, ch. 1] the ideal [N R : M ] is a maximal ideal of R. Since M is coprime, then [N R : M ] = R ann M and so R ann M is a maximal ideal of R, so R = R / R ann M is a field. Thus R is a coprime ring. ■ We note that the condition J(M )  M in prop osition (23) is necessary for examp le: Q as Z-module is coprime module, which has no maximal submodule; that is J(Q) = Q and Z is not a field. Recall that an R-module M is called divisi ble if for each non zero divisor r of R, rM =M ,see[22]. The following p rop osition is an immediate result by theorem (7) and [16, p rop 1.1.7]. Proposi tion (24): Let M be an R-module, then the following st atements are equivalent: 1. M is a cop rime R-module. 2. M is a divisible R / R ann M -module. 3. r M = M for every r  R \ R ann M . 4. I M = M for every ideal I  R ann M . 5. W(M ) = R ann M . Not e that st atement (5) in prop osition (24) was given as definition of coprime module by Abuihlial J. in [7].He defined that an R-submodule N of M is a cop rime submodule if for each homothety r * on M / N is either zero or surjective. Then it is clear that M is a coprime R- module if and only if (0) is a coprime submodule. By combining this result with theorem (7) we have the following corollary . Corollary (25): Let M be an R-module, then the following st atements are equivalent: 1. M is a cop rime R-module. 2. M is a second submodule of itself. 3. (0) is a cop rime submodule of M . The following remark is clear. Remark (26): Let N be a p rop er submodule of an R-module M . Then N is a coprime submodule of M iff M / N is a cop rime R-module. The following result follows by Remark (26) and Not e (8). Corollary (27): Let N be a p rop er submodule of an R-module M , if N is a coprime submodule then R ann (M / N) is a prime ideal. IHJPAS IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VO L.23 (3 ) 2010 Recall that an R-module M is called divisibl e if for each non zero divisor r of R, rM =M , see [22]. Now, we list some consequences of p rop osition (24). Corollary (28): Every faithful cop rime R-module is divisible. Proof: Is clear. ■ Remark (29): Every divisible module over an integral domain is faithful coprime. By combining corollary (28) and remark (29) we get: Corollary (30): Let M be a module over an integral domain R. Then M is a faithful coprime R-module iff M is a divisible R-module. Recall that an R-module M is called principally injective if for every p rincipal ideal I of R and every monomorp hism f: I  M , there is a homomorphism g: R  M such that g / I = f, see [23]. Proposi tion (31): Let M be a module over an integral domain R. Then the following st atements are equivalent: 1. M is a faithful coprime R-module. 2. M is a divisible R-module. 3. M is a principally injective R-module. Proof: (1)  (2) follows by corollary (30). (2)  (3) follows by [12, Exc. 9(a), P.104]. Recall that an R-module M is said to be injective if and only if for any monomorp hism f: A  B where A and B are any two R-modules and for any homomorp hism g: A  M there exists a homomorp hism h: B  M such that h ○ f = g, see [24, p.28]. Corollary (32): If R is PID and M is an R-module, then the following st atements are equivalent: 1. M is a faithful coprime R-module. 2. M is a divisible R-module. 3. M is an injective R-module. Proof: (1)  (2) follows by corollary (.30). (2)  (3). By [26, Th 2.8, P.35]. ■ Corollary (33): Let M be an injective module over an integral domain, then M is coprime. Proof: Since every injective R-module is divisible by [26, Theorem 2.6, p . 33]. Hence the result obt ained by Remark (29). ■ Recall that a module M over an integral domain is called torsion free if (M ) = 0, where (M ) = {m  M ;  r  R, r  0; r m = 0}, see [19, p .45]. Under the class of torsion free modules over an integral domain, we have the following result. Proposi tion (34): Let M be torsion free over an integral domain R. Then the following st atements are equivalent: 1. M is a cop rime R-module. 2. M is a divisible R-module. 3. M is an injective R-module. Proof: (1)  (2). Since M is torsion free, then M is faithful. Thus the result follows by corollary (30). (2)  (3) follows by [26, Th 2.7, p .34]. ■ IHJPAS IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VO L.23 (3 ) 2010 Recall that an integral domain R is called a Dedekind domain if every non-zero ideal of R is invertible and an ideal I is called invertibl e, when I – 1 = {x  Rs: x I  R} and S is the set of non-zero divisors of R, t hen I – 1  I = R, see [25]. Next, we p rove the following. Proposi tion (35): If R is a Dedekind domain and M is an R-module. Then the following st atements are equivalent: 1. M is a faithful coprime R-module. 2. M is a divisible R-module. 3. M is an injective R-module. Proof: (1)  (2) follows by corollary (30). (2)  (3) follows by [24, Prop . 2.10, p .36]. ■ Yassemi S. in [5] introduced the following result without p roof. We give its p roof for sake of completeness. The orem (36): Let M be a p rime R-module. The following st atements are equivalent: 1. M is a cop rime R-module. 2. M is an injective R/ R ann M -module. Proof: Since M is a prime R-module, then by [2, Rem. and Ex. (1.1.3 (3)], R ann M is a p rime ideal, so R is an integral domain and M is a torsion free R = R / R ann M -module, see [20], [11]. (1)  (2). Since M is a cop rime R-module, then by corollary (10) M is a cop rime R - module. Hence by p rop osition (34) M is an injective R -module. (2)  (1). If M is an injective R -module, then by corollary (33) M is a coprime R - module. Hence by corollary (10) M is a cop rime R-module. ■ Recall that an R-module M is called flat if n k k =1 k a = 0, where k  R, ka  M , then there exist b1, …, bn  M and {u i k}  R, i = 1, …, n, k = 1,2,…,r such that n i k i =1 i u b = ka and r i k k =1 k u  = 0, see [2]. The following theorem app eared in [18], we give the details of the p roof for comp leteness. Recall that a subset A of an R-module M is called a basis of M , if A generates M and A is R-linearly indep endent. M is said to be a free R-module, if M has a basis [25, p .190]. The orem (37): Let M be a coprime R-module. Then the following st atements are equivalent: 1. M is a prime R-module. 2. -M is a flat R -module, where R = R / R ann M . Proof: Since M is a coprime R-module, then R = R/ R ann M is an integral domain. Also by p rop osition (24), M is a divisible R -module. IHJPAS IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VO L.23 (3 ) 2010 (1)  (2). Since M is a p rime R-module, then M is a torsion free R -module, where R = R / R ann M . Now, we can show that M is a vector sp ace over K = the total quotient field of R as follows: Let + ann M s + ann M r  K, where r, s  R, s  R ann M . Let m  M , m = (s + R ann M ) m, for some m M since M is a divisible R -module. It follows that + ann M s + ann M r  m = (r + R ann M ) m = r m  M . Thus M is a vector sp ace over K, so it has a basis. It follows that M is a free K-module, hence M is a flat K-module [25, Prop 1.26, p .22]. Then K is a flat R -module by [27, Ex. 20, p .319]. Thus by [19, Exc. 9(a), p .32] M is flat R -module. (2)  (1). [The proof is different from the proof given in [29]. Since R is an integral domain, ( 0 ) is a p rime ideal of R . Hence by [30, Corollary 4.9, ch.1] ( 0 ) M = (0) is a prime R -submodule of M . Thus (0) is a prime R -submodule. Then it is easy to check that (0) is a prime R-submodule of M and hence a prime R-module. Finally, we turn our att ention to the localization of coprime modules. First we have. Proposi tion (38): Let M be a coprime R-module, then M S is a coprime R S -module, where S is a multip licatively closed subset of R. Proof: It follows by theorem () and [16, p rop 1.1.20]. We notice that the converse of p rop osition (38) is not true as the following examp le shows: Let M be Z-module Z and S = Z – {0}, it is clear that S is a multiplicatively closed subset of Z . Z S = Q and Q as Q-module is coprime, but Z is not coprime Z-module. The following corollary follows immediately from prop osition (38). Corollary (39): Let M be a coprime R-module,then M p is a cop rime Rp-module for any p rime ideal P of R. Next, we have the following result. Corollary (40): Let M be a finitely generated R-module. Let S be a multiplicatively closed subset of R. If M is P-cop rime R-module, then M S is PS coprime RS –module. Proof: Since M is P-cop rime, then M is coprime with R ann M = P.T herefore by p rop osition (2.1.38), M S is coprime RS –module. Now, R ann M = [0 R : M ] = P, that is [0 R : M ] S = PS and by [28, p .152], [0S: M S] = PS. Thus SR ann M S = PS. 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IHJPAS 2010) 3( 23مجلة ابن الھیثم للعلوم الصرفة والتطبیقیة المجلد المفھوم الردیف للمقاسات االولیة رشا ابراھیم خلف ، انعام محمد علي جامعة بغداد ، ابن الھیثم –كلیة التربیة ، قسم الریاضیات خالصةال ؛ R لتكن دالیة ذات محاید إذا كان ) د للمقاس األولي مضا كمفھوم(أولیا مضادا مقاسا Mسمى یR.على مقاسا Mحلقة إب M تآلف R= تآلف R   . M في N لكل مقاس جزئي فعلي لھذا المفھوم وكذلك أعطینا عددا ساسیة في ھذا البحث درسنا المقاسات األولیة المضادة ةأعطینا عددا من الخواص األ ن الممیزات لھ تحت أصناف معینة من المقاسات م IHJPAS